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Difference between revisions of "1989 AIME Problems/Problem 14"

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:The only solution to that is <math>(a_1, a_2, a_3)=(4,5,1)</math>.
 
:The only solution to that is <math>(a_1, a_2, a_3)=(4,5,1)</math>.
  
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So we hafour-digitr digit integers <math>(292a_0)_{-3+i}</math> and <math>(154a_0)_{-3+i}</math>, and we need to find the sum of all integers <math>k</math> that can be expressed by one of those.
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So we have four-digit integers <math>(292a_0)_{-3+i}</math> and <math>(154a_0)_{-3+i}</math>, and we need to find the sum of all integers <math>k</math> that can be expressed by one of those.
  
 
<math>(292a_0)_{-3+i}</math>:
 
<math>(292a_0)_{-3+i}</math>:
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We plug the first three digits into base 10 to get <math>10+a_0</math>. The sum of the integers <math>k</math> in that form is <math>145</math>. The answer is <math>345+145=\boxed{490}</math>.
 
We plug the first three digits into base 10 to get <math>10+a_0</math>. The sum of the integers <math>k</math> in that form is <math>145</math>. The answer is <math>345+145=\boxed{490}</math>.
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~minor edit by Yiyj1
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~minor edit by [[User:Yiyj1|Yiyj1]]
  
 
== See also ==
 
== See also ==

Latest revision as of 00:37, 25 August 2024

Problem

Given a positive integer $n$, it can be shown that every complex number of the form $r+si$, where $r$ and $s$ are integers, can be uniquely expressed in the base $-n+i$ using the integers $0,1,2,\ldots,n^2$ as digits. That is, the equation

$r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0$

is true for a unique choice of non-negative integer $m$ and digits $a_0,a_1,\ldots,a_m$ chosen from the set $\{0,1,2,\ldots,n^2\}$, with $a_m\ne 0$. We write

$r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}$

to denote the base $-n+i$ expansion of $r+si$. There are only finitely many integers $k+0i$ that have four-digit expansions

$k=(a_3a_2a_1a_0)_{-3+i}~~$

$~~a_3\ne 0.$

Find the sum of all such $k$,

Solution

First, we find the first three powers of $-3+i$:

$(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i$

So we solve the diophantine equation $a_1-6a_2+26a_3=0 \Longrightarrow a_1-6a_2=-26a_3$.

The minimum the left-hand side can go is -54, so $1\leq a_3 \leq 2$ since $a_3$ can't equal 0, so we try cases:

  • Case 1: $a_3=2$
The only solution to that is $(a_1, a_2, a_3)=(2,9,2)$.
  • Case 2: $a_3=1$
The only solution to that is $(a_1, a_2, a_3)=(4,5,1)$.

So we have four-digit integers $(292a_0)_{-3+i}$ and $(154a_0)_{-3+i}$, and we need to find the sum of all integers $k$ that can be expressed by one of those.

$(292a_0)_{-3+i}$:

We plug the first three digits into base 10 to get $30+a_0$. The sum of the integers $k$ in that form is $345$.

$(154a_0)_{-3+i}$:

We plug the first three digits into base 10 to get $10+a_0$. The sum of the integers $k$ in that form is $145$. The answer is $345+145=\boxed{490}$. ~minor edit by Yiyj1

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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